Is ${404481}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {404481}= &&{4}\cdot100000+ \\&&{0}\cdot10000+ \\&&{4}\cdot1000+ \\&&{4}\cdot100+ \\&&{8}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {404481}= &&{4}(99999+1)+ \\&&{0}(9999+1)+ \\&&{4}(999+1)+ \\&&{4}(99+1)+ \\&&{8}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {404481}= &&\gray{4\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {4}+{0}+{4}+{4}+{8}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${404481}$ is divisible by $3$ if ${ 4}+{0}+{4}+{4}+{8}+{1}$ is divisible by $3$ Add the digits of ${404481}$ $ {4}+{0}+{4}+{4}+{8}+{1} = {21} $ If ${21}$ is divisible by $3$ , then ${404481}$ must also be divisible by $3$ ${21}$ is divisible by $3$, therefore ${404481}$ must also be divisible by $3$.